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November 23, 2014

November 23, 2014

Posted by **Vico** on Thursday, December 3, 2009 at 8:46am.

A person working out in a gym has a metabolic rate of 2500 kJ/h. His body temperature is 37 C, and the outside temperature 24 C. Assume the skin has an area of 2.0 m2 and emissivity of 0.97. (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates 0.40 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) At what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Can someone help and show how to do it?

- Physics -
**drwls**, Thursday, December 3, 2009 at 9:58am(a) Radiative heat loss rate =

(emissivity)(sigma)*(body area)*[T2^4 - T1^4]

T2 is the body temp in Kelvin and T1 is the average temp of surrounding walls.

Sigma is the Stefan-Boltzmann constant. Look it up. Emissivity = 0.97

(b) Multiply the perspiration rate by the heat of skin evaporation

(c) Do the same using the rate of lung moisture evaporation.

(d) Subtract the sum of the previous three mechanisms from 80% of the metabolic rate, which would be 2000 kJ/h.

Express all four in units of kJ/h, for easy comparison.

- Physics -
**Anonymous**, Wednesday, November 12, 2014 at 9:09pm183.06

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